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Suppose we have $n$ planes $H_1, \ldots, H_n$ in $\mathbb{R}^m$ of codimension $q$, or equivalently of dimension $d=m-q$. I want to choose a vector which does not belong to the planes in a continuous way. There are two versions of this problem, depending on how we parameterize the planes, and the answer can be actually different.

  1. Unframed version. Let $Gr_q(m) $ be the grassmannian, that is the space of codimension q planes in $\mathbb{R}^m$. Does there exist a function

$$c : Gr_q(m) ^n \to \mathbb{R}^m$$

Such that $c(H_1, \ldots, H_n) \not \in H_i$ for all $i$?

  1. Framed version. Let $V_{d,m}$ be the Stiefel manifold, that is the space of orthonormal systems in $\mathbb{R}^m$ of cardinality $d$. Does there exist a function

$$c : V_{d,m} ^n \to \mathbb{R}^m$$

Such that $c(H_1, \ldots, H_n) \not \in H_i$ for all $i$?

Note. I slightly changed the notation to agree with Chris one; now $d$ denote the dimension of the planes and $q$ the codimension.

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  • $\begingroup$ Let me also remark that I probably have a proof of this, but it is overcomplicated... It is a question that raised when I was studying configuration spaces, and I am interested in knowing if there is a direct proof like "consider this invariant" or something like that. $\endgroup$ Mar 27, 2021 at 14:30
  • $\begingroup$ What if we relax the codimensione condition, so that the space is not disconnected anymore? $\endgroup$ Mar 27, 2021 at 14:40
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    $\begingroup$ Sorry, I deleted my comment because I thought I had fundamentally misunderstood your question, but I only trivially misunderstood the question! If you reduce the dimension and enforce a codimension of strictly greater than $1$, then I imagine that such a function probably exists, but it likely won't be pretty. $\endgroup$ Mar 27, 2021 at 14:43
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    $\begingroup$ Let us think to the case $d=2, m=3$, that is lines in the space. You can equivalently formulate the problem as: given $n$ points in the projective space $\mathbb{P}^2(\mathbb{R})$, you want to find an $(n+1)$-th point which is distinct from the others. I don't see how this is trivial, even if in this case we relaxed the codimension constraint... To be honest, I think it cannot exist at all! $\endgroup$ Mar 27, 2021 at 14:49
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    $\begingroup$ In the case $d=1$ it's trivial if you use oriented hyperplanes, but false with non-oriented. This is a nice question, by the way. $\endgroup$ Mar 27, 2021 at 17:49

1 Answer 1

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This is a partial answer, though its not particularly elementary. (EDIT: I am using d to denote the dimension of the planes, rather than the codimension as stated in the question, so things need to be re-indexed). If $2d+1\geq m$, $d\neq m$ then for any $n$, we cannot find such a map. Assume such a map exists, and consider the real steifel manifolds $V_{m,d}$. These are the space of orthonormal $d$ tuples in $\mathbb{R}^n$, and we have a natural surjective map $V_{m,d}\xrightarrow{\pi} V_{m,d-1}$ given by forgetting the last vector in our list. By restricting a hypothetical $c$ map to the first coordinate (with other coordinates some fixed spaces), it suffices to show that no map can exist when $n=1$.

We claim that a map of the form described will yield a section of the map $$V_{m,d+1}\xrightarrow{\pi}V_{m,d}$$

To see this, first normalise the map $c$ to land in unit one vectors in $\mathbb{R}^n$. Then for a given orthonormal $d$ frame, map it to the orthonormal $d+1$ frame with new vector $c(Span(v_1,..v_{d}))$. This is clearly a section, so now consider mod $2$ cohomology, and in particular the cup product structure.

By results of Borel we know this graded ring, which I read at https://www.researchgate.net/publication/254207624_The_cohomology_rings_of_real_Stiefel_manifolds_with_integer_coefficients

The cohomology ring of $V_{m,d}$ is generated by classes of generators $z_i$ of degree $i$ for $m-d\leq i\leq m-1$, subject to the relation $z_i^2=z_{2i}$ when $2i\leq m-1$, and $z_i^2=0$ else. The map $\pi$ induces the natural inclusion of these rings, so consider $z_{m-d-1}^2=z_{2m-2d-2}$ in $H^*(V_{m,d+1},\mathbb{Z}/2\mathbb{Z})$. On one hand, our section is the identity on this element, since $z_{2m-2d-2}$ is in the image of the natural inclusion, but on the other, the minimal positive degree nonzero cohomology group of $V_{m,d}$ is in degree $m-d$, so $z_{m-d-1}$ must be in the kernel. So since the induced map on cohomology rings is a map of graded rings, no such map $c$ can exist in this case.

On the other side, the case of $d=1$, $m> 2$ is also not possible. Similarly, reduce to the $n=1$ case, and thus we obtain a map $\mathbb{RP}^{m-1}\rightarrow S^{d-1}$ such that when we compose with the natural quotient $S^{d-1}\rightarrow \mathbb{RP}^{m-1}$, the induced endomorphism of $\mathbb{RP}^{m-1}$ has no fixed points. Then consider the trace of this endomorphism on mod $2$ cohomology. Since it has no fixed points, this trace is $0$, and thus this endomorphism is nonzero on some nonzero cohomology group (since its an isomorphism on $H^0$), so induces an isomorphism on $H^{m-1}(\mathbb{RP}^{m-1},\mathbb{Z}/2\mathbb{Z})$, since all the other maps factor through $0$. But a generator for this top cohomology is the $m-1$ fold cup product of the class in $H^1$, so this must be the zero map (since $S^{m-1}$ is simply connected). Thus, no map can exist in this case either.

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  • $\begingroup$ Hi Chris H, thank you for your interesting points. Regarding the second case, you are formally right, though this can be solved by considering the oriented hyperplanes. In this case the $n=1$ case is possible, while the $n\ge 2$ is impossible because more than one hyperplane disconnects the space. Regarding the first obstruction, in principle $c(Span(v_1, \ldots, v_d)) $ is not orthogonal to $v_1, \ldots, v_d$, so you have to graham-schmidt somehow. Does this process preserve continuity? $\endgroup$ Mar 28, 2021 at 7:52
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    $\begingroup$ Ahh yep, that’s my bad, one does indeed need to use GS to orthogonalise, but this is fine, this a continuous process, the algorithm is just additions/scalar multiplications. One can also avoid it too, if one works with the noncompact steifel manifold, with no orthonormality relations imposed. The continuity of GS yields that these spaces are homotopy equivalent. $\endgroup$
    – Chris H
    Mar 28, 2021 at 8:12
  • $\begingroup$ There is also a little misunderstanding on the $d$: in your solution is the dimension of the planes, while I stated d as the codimension. I think you inequality becomes then $2d-1 \le m$. And also $d > 0$, otherwise the problem is trivial impossible. I like your formulation with stiefel manifolds because parameterize planes with an ortonormale basis. This prevents non-geometric obstruction given by monodromy, like the one you found in the second paragraph. I will add this "framed version" to the question. $\endgroup$ Mar 28, 2021 at 8:15
  • $\begingroup$ I edited to address the indexing issue, and one interesting facet of this question is whether this problem is always reducible to the n=1 case, that is, if the obstruction ever actually depends on n. $\endgroup$
    – Chris H
    Mar 28, 2021 at 8:28
  • $\begingroup$ The framed version in codimension one depends on whether $n\ge 2$ or $n=1$. In other cases, I actually agree that since the things are "thin" and we have excluded monodromy, the problem that can arise is that we cannot add a new "canonical" point to some space (as you demonstrated, and I still can't understand what is really going on). Let's see if someone other can settle the $n=1$ case by looking at the fibration you mentioned! $\endgroup$ Mar 28, 2021 at 11:22

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